# Composite functions

Composite functions: Let A, B, C are three sets .Let f: A -> B, g : B -> C be two functions. Here we have taken the domain of g to be the

co-domain of f.

g o f : A -> C as

g o f (a) = g [f (a) ] for a ∈ A

Since f(a) ∈ B

g [f(a) ] ∈ C

The function g o f so obtained is called the composition of f and g.

## Examples on composite functions

**Example 1:**A = {1, 2, 3}, B = {4, 5}, C = {5, 6}

Let f : A -> B, g: B -> C be defined by f(1) = 4, f(2) = 5, f(3) = 4, g(4) = 5, g(5) = 6. Find g o f : A -> C

**Solution:**We have,

f(1) = 4 and g (4) = 5

So g [f(1) ] = g o f (1) = 5

f(2) = 5 and g(f) = 6

∴ g [f(2)] = g o f (2) = 6

f(3) = 4 and g(4) = 5

So g [f(3)] = g o f (4) = 5

So, g o f = {(1, 5), (2, 6), (3, 5)}

**Example 2:**Let f:R -> R be defined by f(x) = 2x – 3 and g:R -> R be defined by g(x) = (x + 3)/ 2. Show that f o g = g o f

**Solution :**f o g is a composite-function.

f o g means g(x) function is in f(x) function.

f o g = f[g(x)]

f(x) = 2x -3 and g(x)= (x+3)/2

∴ f[g(x)] =[2(x+3)/2] - 3

= x + 3 - 3

f o g = x -----(1)

g o f means f(x) function is in g(x) function.

g o f = g[f(x)]

f(x) = 2x -3 and g(x) = (x+3)/2

g[f(x)] = [(2x-3) + 3]/2

= (2x -3 + 3)/2

= 2x/2

g o f = x ----(2)

So from equation (1) and (2)

f o g = g o f

**Example 3:**Let f, g: R -> R be defined respectively, by f(x) = x

^{2}+ 3x + 1 , g(x) = 2x -3. Find f o g (2).

**solution :**f o g means g(x) function is in f(x) function.

This means put x = 2x -3 in f(x) function.

f[g(x)] = (2x - 3)

^{2}+ 3 (2x -3)+ 1

= (2x)

^{2}- 2(2x)(3) + 3

^{2}+ 6x - 9 + 1

= 4x

^{2}- 12x + 9 + 6x - 9 +1

f[g(x)] = f o g (x)= 4x

^{2}-6x + 1

Now we have to find f o g(2), so put x = 2 in f o g

f o g (2)= 4(2)

^{2}- 6(2) +1

= 4 x 4 - 12 + 1

= 16 -12 +1

∴ f o g(2) = 5

**11th grade math**

From Composite functions to Home page

From Composite functions to Home page